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Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(VT=\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{bd.k^2}{bd}=k^2\left(1\right)\)
\(VP=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)
cho \(b^2=a.c-a^2=b.d\)
c/m:\(\dfrac{a^3+b^3-c^3}{b^3+c^2-d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^2\)
\(b^2=a.c\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\left(1\right)\)
\(c^2=b.d\)
\(\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b-c}{b+c-d}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{b}{c}\right)^3=\left(\dfrac{c}{d}\right)^3=\left(\dfrac{a+b-c}{b+c-d}\right)^3\)
\(=\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3=\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}\left(đpcm\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk\),\(c=dk\)
\(\dfrac{a^2}{b^2}=\dfrac{bk^2}{b^2}=k^2\left(1\right)\)
\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\left(2\right)\)
Từ (1) và (2)=>\(\dfrac{a^2}{b^2}=\dfrac{ac}{bd}\)(đpcm)
Đặt \(\dfrac{a}{b}=k;\dfrac{c}{d}=k\)
\(\Rightarrow a=kb;c=kd\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{bk^2}{b^2}=k^2\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bkdk}{bd}=k^2\)
Từ các chứng minh trên cho ta thấy
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{a.c}{b.d}\)
+ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
+ \(\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\) \(\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
+ \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a^2}{c^2}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(\Rightarrow\frac{a\cdot b}{c\cdot d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
\(\Rightarrow\frac{a}{b}\cdot\frac{a}{b}=\frac{a^2+c^2}{b^2+d^2}\Rightarrow\frac{a\cdot c}{b\cdot d}=\frac{a^2+c^2}{b^2+d^2}\)
câu cuối lm tương tự
Ta có \(\frac{a}{b}=\frac{c}{d}=>\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=>\frac{a}{a-b}=\frac{c}{c-d} \)
1. Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) \(\left(1\right)\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) \(\left(2\right)\)
Từ \(\left(1\right)\text{và (2)}\) \(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)
2. \(\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
\(\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|\ge0\\\left|\dfrac{2}{7}y+3\right|\ge0\end{matrix}\right.\Rightarrow\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|\ge0\)
\(\text{Mà }\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|=0\\\left|\dfrac{2}{7}y+3\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5-\dfrac{3}{4}x=0\\\dfrac{2}{7}y+3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{4}x=5\\\dfrac{2}{7}x=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{3}\\y=-\dfrac{21}{2}\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}x=\dfrac{20}{3}\\y=-\dfrac{21}{2}\end{matrix}\right.\)
3. \(\dfrac{1}{2}a=\dfrac{2}{3}b=\dfrac{3}{4}c\)
\(\Rightarrow\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}\)
\(\text{Mà }a-b=15\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}=\dfrac{a-b}{2-\dfrac{3}{2}}=\dfrac{15}{\dfrac{1}{2}}=30\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=30\Rightarrow a=30.2=60\\\dfrac{b}{\dfrac{3}{2}}=30\Rightarrow b=30.\dfrac{3}{2}=45\\\dfrac{c}{\dfrac{4}{3}}=30\Rightarrow c=30.\dfrac{4}{3}=40\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}a=60\\b=45\\c=40\end{matrix}\right.\)
Đề sai rồi bạn. Phải thay "^2" bằng "^3" mới đúng.
xinloi mik ghi nhầm đề bài